The present invention provides a pipe processing device for cutting of tubes and pipes of uniform cross section and more particularly, to a process having two blades to cut the tubes or pipes wherein one is approaching the tube/pipe from the top and the other from the bottom. Further, the invention pertains to a device with 2 blades to cut tubes of high OD and thickness increasing productivity of the production line by 50 %. The present device automatically optimizes the trajectory of the 2 blades in order to cut the tubes in minimum time and with maximum accuracy. Moreover the accuracy in cut length is within 1 mm.
Background of the invention
In the online tube forming process, cutting tubes at desired length makes for a very crucial operation. The speed of the entire production line depends on this process and therefore executing this process effectively (avoiding accidents) is of crucial importance.
Tubes are cut using a rotating circular saw blade. As the diameter of the tube increases, the distance which the saw blade travels to execute the cut also increases. This increases the cutting time and as a result reduces the speed of the tube production line, lowering productivity and increasing the per unit production cost.

In addition to the above mentioned challenges, there is a possibility that the 2 blades may collide since they are moving in the same plane and axis. Therefore the device has to be intelligently designed so as to account for this possibility and incorporate anti-collision protocols and algorithms.
In order to overcome the above said problems, the present invention provides a pipe processing device and process for cutting of tubes and pipes of uniform cross section.
Objectives of the Invention
This invention is to provide a pipe processing device for use in cutting of tubes/pipes of any diameter.
The object of this invention is to provide the pipe processing device which can cut a pipe/tube by circular saw blades at desired length by increasing the productivity of the production line by 50 %.
Another object of this invention is to provide the pipe processing device in order to reduce the cutting time of the tube/pipe having larger diameter and thickness.
Further object of this invention is to provide the process of cutting a tube/pipe with the said device wherein the length accuracy of the tube cut is within 1 mm. For example if set length is 6000 mm, the tube cut will be between 5999 mm & 6001 mm.

Brief Description of Drawings
The drawings described herein are for illustrative purposes only of selected embodiments and not all possible implementations, and are not intended to limit the scope of the present disclosure.
Figure 1 shows that the cutting plane along the diametric center of the blades is off set from the diametrical center of the tube by 14.608 mm wherein the top blade diameter is 500 mm, bottom blade diameter is 500 mm and tube diameter is 114 mm.
Figure 2 shows that the saw movement line is offset from the tube center by 25 mm above wherein the top blade diameter is 450 mm, bottom blade diameter is 450 mm and tube diameter is 90 mm.
Figure 3 shows that the saw movement line is offset from the tube center by 14.608mm above wherein the top blade diameter is 500 mm, bottom blade diameter is 500 mm and square tube is 100 x 100 mm.
Figure 4 showsthat the cutting plane along the diagonal center of the blades is off set from the diametrical center of the tube by 4.878 mm wherein the top blade diameter is 500 mm, bottom blade diameter is 500 mm and section tube is 100 x 150 mm.
Figure 5 to 16 show the examples of the calculations based on different diameter of tubes and blades.
Detailed description of the invention:
The following is a detailed description of exemplary embodiments to illustrate the principles of the invention. The embodiments are provided to illustrate aspects of the

invention, but the invention is not limited to any embodiment. The scope of the invention encompasses numerous alternatives, modifications and equivalent; it is limited only by the claims.
Numerous specific details are set forth in the following description in order to provide a thorough understanding of the invention. However, the invention may be practiced according to the claims without some or all of these specific details. For the purpose of clarity, technical material that is known in the technical fields related to the invention has not been described in detail so that the invention is not unnecessarily obscured.
An aspect of the present invention is directed towards a pipe processing device and process for cutting of tubes and pipes of uniform cross section.
One embodiment of the present invention provides the device with 2 blades in order to reduce the cutting time of tubes having larger diameter and thickness. One blade approaches the tube from the top and the other blade from the bottom. This reduces the effective cutting time to almost 50% as that obtained using a single blade for cutting.
Tube cutting using 2 blades is a complicated process and has to account for many situations wherein the 2 blades could collide and lead to an accident or a situation where the tube is not fully cut due to insufficient movement of blades. This challenge is further amplified as the diameter of the 2 blades can be different and also the diameter and thickness of the tubes will change based on the batch being manufactured by the user.

It may so happen that the diameters of the 2 blades are not the same due to non-availability of blades with similar diameters and thickness. The arrangement becomes very complicated because tube centre is not in line with saw movement line. This eccentricity increases or decreases depending on the diameter of the tubes as the tubes are always placed on a fixed plate. So, the tube centre sometimes goes above the saw movement line depending on tube diameter. The important aspect is to find out the first contact point and the last contact point between the tube and the saw. It is therefore imperative to ensure that the above parameters are taken into account while executing the cut.
In the present process the above permutations and combinations have been taken into account and the calculations automated in a manner that the user only needs to enter the respective diameters and thickness of the tubes. The cutting process itself is very efficient and executes the trajectory of the blades such that the blades only travel a length of about 1 mm more than the critical path of motion needed to fully execute the cut successfully. The challenges mentioned above are graphically illustrated below for better understanding.
The length which the formulae calculate is the Excess Length which the Top Blade must travel over and above the blade collision point (The point at which the 2 blades would theoretically collide). Also this Excess Length is the length which the bottom blade would stop before the Collision Point. This would ensure that the blades move in a non symmetric fashion - ensuring that the tube is completely cut (factoring the shift in the tube center) with minimum excess movement of the blades. (The extra movement which the blades are doing is within 1 mm of the geometrical most optimal

distance required to be traveled by the blades to execute this cut). For example, please refer figure 1 below: -
Figure 1 illustration: Top Blade Diameter: 500 mm, Bottom Blade Diameter: 500 mm & Tube Diameter: 114 mm. This illustration shows that the cutting plane along the diametric center of the blades is off set from the diametrical center of the tube by 14.608 mm.
In this the saw movement line is offset from the tube center by 14.608 mm above. The extra distance one saw needs to move is 20.403 mm to complete the cutting and the other saw should move less by 20.403mm from the virtual contact point of the saws to avoid accident. When both the blades move along the cutting plane, the portion of uncut tubes is 7.202 mm and 20.403 mm respectively. This is significant because it clearly demonstrates that uniform movement of both blades leads the tube to remain uncut and also that 1 blade has to travel beyond the center point of the tube. It is interesting to note at this point that quantifying the extra movement of either of the blades is pertinent to ensure the cutting time remains minimal.
Figure 2 illustration: Top Blade Diameter: 450 mm, Bottom Blade Diameter: 450mm & Tube Diameter: 90 mm. This illustration shows that the saw movement line is offset from the tube center by 25 mm above. The extra distance one saw needs to move is 25 mm to complete the cutting and the other saw should move less by 25 mm from the virtual contact point of the saws to avoid accident. When both the blades move along the cutting plane, the portion of uncut tubes is 17.78 mm and 21.489mm

respectively. This is significant because it clearly demonstrates that uniform movement of both blades leads the tube to remain uncut and also that 1 blade has to travel beyond the center point of the tube. It is interesting to note at this point that quantifying the extra movement of either of the blades is pertinent to ensure the cutting time remains minimal.
Figure 3 illustration: Top Blade Diameter: 500 mm, Bottom Blade Diameter: 500 mm & Square Tube: 100x100mm. This illustration shows that the saw movement line is offset from the tube center by 14.608mm above. The extra distance one saw needs to move is 20.403 mm to complete the cutting and the other saw should move less by 20.403mm from the virtual contact point of the saws to avoid accident.
When both the blades move along the cutting plane, the portion of uncut tubes is 11.275 mm and 21.654 mm respectively. This is significant because it clearly demonstrates that uniform movement of both blades lead the tube to remain uncut and also that 1 blade has to travel beyond the center point of the tube. It is interesting to note at this point that quantifying the extra movement of either of the blades is pertinent to ensure the cutting remains minimal so as to not adversely impact the cutting time.
Figure 4 illustration: Top Blade Diameter: 500 mm, Bottom Blade Diameter: 500 mm & Section Tube: 100x150mm. This illustration shows that the cutting plane along the diagonal center of the blades is off set from the diametrical center of the tube by 4.878 mm. When both the blades move along the cutting plane, the portion of uncut tubes is

18.945 mm and 13.317 mm respectively. This is significant because it clearly demonstrates that uniform movement of both blades lead the tube to remain uncut and also that 1 blade has to travel beyond the center point of the tube. It is interesting to note at this point that quantifying the extra movement of either of the blades is pertinent to ensure the cutting time remains minimal.
Now calculating the uncut length by using standard geometric/ trigonometric formulae was quite complicated. The present invention provides equations for different cases which satisfied the result within 1 mm in different combinations.
Dl =tube diameter
D2 = blade diameter
S = offset
(AL)B = BOTTOM RC LENGTH = Di°36 x D2° 2 x S° 034 x (500/D2)
When tube diameter is more than 148mm, means when tube center comes above blade
center line then following equations to be used.
(AL)B = BOTTOM RC LENGTH = Di°35 x D2° 2 x S° 034 x (500/D2)
For square tube
D = Length of one side of the square
Di = 4D/TI
B = Di036x D202 xS0034 X (500/D2) + 2
For rectangular tube keeping smaller side vertical
B = Di0276 XD202X S0034X (500/D2)

Figure 5 to 16 show the examples of the calculations based on different diameter of tubes and blades.
In reference of Figure 5, the example represents for tube with 114 diameter and blade with 500 diameter. The calculation is as follows:
114 diameter tube and 500 diameter blade
(CL)B = Di036x D202 x S0034x (500/D2)
S = [28.868-(Di/2 - 45)] x 0.866
= [28.868-Q 14/2 - 45)] x 0.866
= [28.868-(57-45)] x 0.866
= [28.868- 12] x 0.866 = 16.868x0.866 = 14.608
(CL)B = (114)036x (500)02 x (14.608)0034x (500/500)
=5.50155 x 3.4657 x 1.09545 x 1
= 20.8868mm
Actual available value is 20.403, hence the difference is even less than 1mm.
In reference of Figure 6, the example represents for tube with 127 diameter and blade with 500 diameter. The calculation is as follows:
127 diameter tube and 500 diameter blade
(CL)B = Di036x D202 x S0034x (500/D2)
S = [28.868 - (Di/2 - 45)] x 0.866
= [28.868 - (127/2 - 45)] x 0.866
= [28.868 - (63.5 - 45)] x 0.866
= [28.868-18.5] x 0.866
= 10.368x0.866

(CL)B = (127)036x (500)°-2X (8.9786)0034X (500/500)
=5.7196 x 3.4657 x 1.0775x1
= 21.3585mm
Actual available value is 20.951; hence the difference is even less than 1mm.
In reference of Figure 7, the example represents for tube with 168 diameter and blade
with 500 diameter. The calculation is as follows:
168 diameter tube and 500 diameter blade, in this case tube center in above the saw movement line.
S = [28.868 - (Di/2 - 45)] x 0.866
= [28.868 - (168/2 - 45)] x 0.866
= [28.868-(84-45)] x 0.866
= - 8.7743
Minus sign indicates that the tube center in above the saw movement line
(CL)B = Di036x D202 x S0034x (500/D2)
= (168)036x (500)02x (8.7743)0034x (500/500)
= 6.32574 x 3.4657 x 1.0766
= 23.61555
A= 96.9768 (90/Di)0064 x D20087
= 96.9768 (90/168)0064 x 5000087
= 96.9768x0.96084x1.7171
= 160
AL = (360 - 2A) x (7C x Di / 360)
= (360 - 2 x 160) x (71 x 168/ 360)
= 40XTIX 168/360
= 58.643
(AL)T = AL-(AL)B
= 58.643-23.61555

= 35.0275
Actual available value is 34.305; hence the difference is even less than 1mm.
When the blades are of different diameter, one blade is 500 diameters as another blade
of 450mm diameter.
In reference of Figure 8, the example represents for tube with 100 x 100 (127(|)) diameter and blade with 500 diameter. The calculation is as follows:
100 x 100 (127<|>) diameter tube and 500 diameter blade D = Length of one side of square
Di = 4D/ 7i = 4 x 100/ 7C = 127.323
D2 = Blade diameter = 500 S = [28.868 - (Di/2 - 45)] x 0.866 = [28.868 - (127/2 - 45)] x 0.866
= [28.868-18.5] x 0.866 = 8.9786
(CL)B = Di036x D202 x S0034x (500/D2)
= (127)036x (500)02 x (8.979)0034x (500/500)
= 5.7196 x 3.4657 x 1.0775x1
= 21.3585
Actual available value is 21.658; hence the difference is even less than 1mm.
In reference of Figure 9, the example represents for tube with 125 x 125 (159 did) and blade with 500 diameter. The calculation is as follows:
125 x 125 (159 did) tube and 500 diameter blade

D = Length of one side of square
Di = 4D/7i = 4 x 125/7r= 159.15 D2 = Blade diameter 500mm
S = [28.868 - (Di/2 - 45)] x 0.866
= [28.868 - (159/2 - 45)] x 0.866
= [28.868-79.5-45] x 0.866
= (28.868-34.5) x 0.866
= -4.877 mm
Minus sign indicates that the tube center in above the blade movement line
(CL)T= 1.33DI°-35X D202 x S0034x (500/D2)
= 1.33 x 159035x(500)°-2x(4.877)0034x (500/500)
= 1.33 x 5.895 x 3.4657 x 1.0554 x 1
= 28.6767
Actual available value is 29.457;hence the difference is even less than 1mm.
In reference of Figure 10, the example represents for tube with 145 x 82 (144.5$) diameter and blade with 500 diameter. The calculation is as follows:
For rectangle tube keeping smaller side vertical
145 x 82 (144.5(|))Tube and 500 diameter blade, Di is the length of the smaller side of the rectangle
D = Length of bigger side of rectangle d = Length of smaller side of rectangle
Di = 2 (D+d)/ 7C = 2 (145+82)/ % = 2 x 227/TC = 554/TC = 144.512 s 144.5

S = [28.868 - (Di/2 - 45)] x 0.866
= [28.868 - (144.5/2 - 45)] x 0.866
= [28.868-27.25] x 0.866
= 1.4011
(CL)B = DI°-276X D20-2 x S0034x (500/D2)
= 820276x (500)02x (1.401 l)0034x (500/500)
= 3.3745 x 3.4657 x 1.01153 x 1
= 11.83
Actual available value is 11.349; hence the difference is even less than 1mm.
In reference of Figure 11, the example represents for rectangle tube keeping bigger side vertical. The calculation is as follows:
For rectangle tube keeping bigger side vertical
(CL)B = Length of the smaller side/5 + 6 = 82/5 + 6 = 16.4 + 6 = 22.4 mm
Figure 12 represents the example in respect to the figure 5, 114 diameter tube and 500 diameter blade at the top on 450 diameter blade at the bottom. Calculated value (example 1) 20.8868 as actual value is 21.526, hence the difference is less than 1 mm.
Figure 13 represents the example in respect to the figure 6, 127 diameter tube and 500 diameter blade at the bottom and 450 diameter blade at the top. Calculated value (example 2) is 21.3585 as actual available value is 22.107; hence the difference is less than 1 mm.

Figure 14 represents 168 diameter tube and 500 diameter blade at the bottom and 450 diameter blade at the top 168 diameter tube. Calculated value is 35.03 mm and actual available value is 36.196, hence the difference is 1.166 mm only.
In reference of Figure 15, the example represents forlOO x 100 (127 diameter) tube and 500 diameter blade at the bottom and 450 diameter blade at the top . The calculation is as follows:
100 x 100 (127 diameter) tube and 500 diameter blade at the bottom and 450 diameter blade at the top
D 1=4x1 00/TI= 400/7i= 127.323 == 127 Blade diameter is considered as 450
s = [28.868 - (Di/2 - 45)] x 0.866 =[28.868 - (117/2 - 45)] x 0.866
= (28.868-18.5) x 0.866
= 8.9786
(CL)B = Di036x D202 x S0034x (500/D2)
= (127)036x (450)02 x 8.9786x (500/450)
= 5.7196x3.3935x8.9786x1.1111
= 23.2369
Actual available values are 22.552 and 23.495 mm for interchanging the position of blades. However the difference is even less than 1mm.
In reference of figure 16, the example represents for 125 x 125 (159 diameter) tube and 500 diameter blade at the bottom and 450 diameter blade at the top. The calculation is as follows:

D1=4D/TI = 4X 125/71= 159.15 s 159 D2 = Blade diameter 500mm.
S = [28.868 - (Di/2 - 45)] x 0.866
= [28.868 - (159/2 - 45)] x 0.866
= [28.868 - (79.5- 45)] x 0.866
= (28.868-34.5) x 0.866
= -4.877 mm
Minus sign indicates that the tube center in above the blade movement line.
(CL)T= 1.33DI°-35X D202 x S0034x (500/D2)
= 1.33 x 159035x (500)02x (4.877)0034x (500/450)
= 1.33 x 5.895 x 3.4657 x 1.0554 x 1.1111
= 31.8637
Actual available values are 30.625 and 32.221 mm for interchanging the position of blades. However the difference is even less than 1mm.
The example is for rectangular tube keeping smaller side vertical
145 x 82 (144.5) tube and 500 diameter blade at the bottom and 450
diameter blade at the top D = Length of bigger side of rectangle d = Length of smaller side of rectangle
Dl=2(D+d)/7i = 2(145+82)/TC = 2x 227/TC = 554/TC = 144.512 s 144.5
S = [28.868 - (Di/2 - 45)] x 0.866 = [28.868 - (144.5/2 - 45)] x 0.866 = (28.868-27.25) x 0.866 = 1.4011

(CL)B= DI0276X D202 x S0034x (500/D2)
= 820276x (500)02x (1.401 l)0034x (500/500)
= 3.3745 x 3.4657 x 1.01153 x 1
= 11.83
Actual available values are 12.099 and 11.773 mm for interchanging the position of blades. However the difference is even less than 1mm
The example is for rectangle tube keeping the bigger side vertical
(CL)B = Length of the smaller side/5 + 6 = 82/5 + 6 = 16.4 + 6 = 22.4 mm
Although preferred embodiments of the invention have been illustrated and described, it will at once be apparent to those skilled in the art that the invention includes advantages and features over and beyond the specific illustrated construction. Accordingly, it is intended that the scope of the invention be limited solely by the scope of the hereinafter appended claims, and not by the foregoing specification, when interpreted in light of the relevant prior art.

WE Claim

1.A pipe processing device for cutting of tubes and pipes of uniform
cross section comprising:
a set of two blades to cut the tubes or pipes wherein one blade is approaching the tube/pipe from the top and the other blade from the bottom in which one blade travels extra length and other travels less by same length to successfully cut the tube and this excess length for tube of different diameters and blades of different diameters is calculated with the formula as:
(AL)B = Bottom RC Length = Di036 x D202 x S0034 x (500/D2)
and when the tube diameter is more than 148mm, means when tube center comes above blade center line then the equation to be used as,
(AL)B = Bottom RC Length = D1035 x D202 x S0034 x (500/D2) in which
Dl =tube diameter D2 = blade diameter and S = offset.
2. The device as claimed in claim 1 wherein the formula for square tube
is as:
B = Dl036 x D202 x S0034 x (500/D2) + 2

3. The device as claimed in claim 1 wherein the formula for rectangular
tube is as:
B = Di0276 x D202 x S0034 x (500/D2)
4. A process for cutting of tubes and pipes of uniform cross section
comprising:
a set of two blades to cut the tubes or pipes wherein one blade is approaching the tube/pipe from the top and the other blade from the bottom in which one blade travels extra length and other travels less by same length to successfully cut the tube and this excess length for tube of different diameters and blades of different diameters is calculated with the formula as:
(AL)B = Bottom RC Length = Di036 x D202 x S0034 x (500/D2)
and when the tube diameter is more than 148mm, means when tube center comes above blade center line then the equation to be used as,
(AL)B = Bottom RC Length = Di035 x D202 x S0034 x (500/D2) in which
DI =tube diameter D2 = blade diameter and S = offset.